Analytic solution of $ \int_t^{t+r} \frac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx $

Question

Could anyone help me to solve this defined integral analytically? $$ \int_t^{t+r} \dfrac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx $$

Answer 1

The antiderivative of the integrand of the integral under consideration can be found by the Euler changes $\sqrt{2rx-x^2}=tx$ or/and $\sqrt{2rx-x^2}=t(x-2r)$. The output done with Maple by the command $int(x^2/(r+t-2*sqrt(2*r*x-x^2))^4, x)$ is huge. It can be seen in the worksheet exported as a pdf file (20MB). The definite integral under consideration depends on the two parameters $t$ and $r$ so parametric analysis is necessary. I give only its results:$ t \le r, t \ge r/3$ under the assumption $t \ge 0$. In other cases the integral may diverge because of the singularity caused by the zeros of the denominator of the integrand $r+t-2\sqrt{2rx-x^2}=0$:$$x_1=r+\frac 1 2\,\sqrt {3\,{r}^{2}-2\,rt-{t}^{2}},x_2=r-\frac 1 2\,\sqrt {3\,{r}^{2}-2\,rt-{t}^{2}}. $$ For concrete values of the parameters the integral under consideration can be found with Maple by the command $$J:=(t,r)->int(x^2/(r+t-2*sqrt(2*r*x-x^2))^4,x=t..t+r).$$ For example, $$J(1,2)= {\frac {1440}{2401}}\,{\it arctanh} \left( 2/7\,\sqrt {7} \right) \sqrt {7}+{\frac {360}{2401}}\,\sqrt {7}\ln \left( \sqrt {7}\sqrt {3} +8\,\sqrt {3}-9 \right) -$$ $${\frac {360}{2401}}\,\sqrt {7}\ln \left( \sqrt {7}\sqrt {3}+8\,\sqrt {3}+9 \right) +$$ $${\frac {360}{2401}}\,\sqrt {7}\ln \left( -\sqrt {7}\sqrt {3}+8\,\sqrt {3}+9 \right) -$$ $${\frac {360 }{2401}}\,\sqrt {7}\ln \left( -\sqrt {7}\sqrt {3}+8\,\sqrt {3}-9 \right) +{\frac {27404}{3087}}\,\sqrt {3}+{\frac {15494}{1029}} $$ and $$J\left(\frac 1 {10},2\right) = \infty.$$

Answer 2

This is not an analytic answer in terms of known functions, but in an attempt to try to understand the complexity of your question I looked at the Taylor series expansion around $x=0$:

Mathematica 8:

 Clear[x, r, t]
 Series[x^2/(r + t - Sqrt[(2 r x - x^2)])^4, {x, 0, 3}]

$$\dfrac{x^2}{(t+r-\sqrt{2rx-x^2})^4} = \frac{x^2}{(r+t)^4}+\frac{4 \sqrt{2} \sqrt{r} x^{5/2}}{(r+t)^5}+\frac{20 r x^3}{(r+t)^6}+\frac{\sqrt{2} \left(39 r^2-2 r t-t^2\right) x^{7/2}}{\sqrt{r} (r+t)^7}+O[x]^4$$

with Big Oh-notation at the end.

Cutting the Big Oh notation at the end and only integrating the four first terms in the Taylor series in Mathematica with limits $t$ and $t + r$:

Integrate[
 x^2/(r + t)^4 + (4 Sqrt[2] Sqrt[r] x^(5/2))/(r + t)^5 + (
  20 r x^3)/(r + t)^6 + (Sqrt[2] (39 r^2 - 2 r t - t^2) x^(7/2))/(
  Sqrt[r] (r + t)^7), {x, t, t + r}]

we get the Conditional expression:

$$\text{ConditionalExpression}\left[\frac{1}{63 (r+t)^7}\left(-t^3 \left(21 r^3+72 \sqrt{2} r^{5/2} \sqrt{t}+378 r^2 t+690 \sqrt{2} r^{3/2} t^{3/2}+378 r t^2+44 \sqrt{2} \sqrt{r} t^{5/2}+21 t^3-\frac{14 \sqrt{2} t^{7/2}}{\sqrt{r}}\right)+\frac{(r+t)^4 \left(336 r^{5/2}+357 r^{3/2} t+21 \sqrt{r} t^2+618 \sqrt{2} r^2 \sqrt{r+t}+44 \sqrt{2} r t \sqrt{r+t}-14 \sqrt{2} t^2 \sqrt{r+t}\right)}{\sqrt{r}}\right),((\text{Im}[r]\leq 0\&\&\text{Im}[r] \text{Re}[t]\leq \text{Im}[t] \text{Re}[r])\|(\text{Im}[r] \text{Re}[t]\geq \text{Im}[t] \text{Re}[r]\&\&\text{Im}[r]\geq 0))\&\&\left(\left(\text{Re}\left[\frac{t}{r}\right]\geq 0\&\&r\neq 0\&\&t\neq 0\right)\left\|\frac{t}{r}\notin \text{Reals}\right\|\text{Re}\left[\frac{t}{r}\right]\leq -1\right)\right]$$

I am not sure this helps though.